exemplar-phy-12-ray-optics-and-optical-instruments

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<p><strong>Ques.</strong> You are given four sources of light each one providing a light of a single colour - red, blue, green and yellow. Suppose the angle of refraction for a beam of yellow light corresponding to a particular angle of incidence at the interface of two media is $90^{\circ}$. Which of the following statements is correct if the source of yellow light is replaced with that of other lights without changing the angle of incidence?</p>
<div class="fragment">
<p>(a) The beam of red light would undergo total internal reflection</p>
<p>(b) The beam of red light would bend towards normal while it gets refracted through the second medium</p>
<p>(c) The beam of blue light would undergo total internal reflection</p>
<p>(d) The beam of green light would bend away from the normal as it gets refracted through the second medium</p>
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<p><strong>Thinking process :</strong></p>
<p>This problem is based on the critical angle of total internal reflection.</p>
<p><strong>Ans.</strong> (c)</p>
<p>According to VIBGYOR, among all given sources of light, the blue light have smallest wavelength.According to Cauchy relationship, smaller the wavelength higher the refractive index and consequently smaller the critical angle.</p>
<p>So, corresponding to blue colour, the critical angle is least which facilitates total internal reflection for the beam of blue light. The beam of green light would also undergo total internal reflection.</p>
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<p><strong>Ques.</strong> The direction of ray of light incident on a concave mirror is shown by PQ while directions in which the ray would travel after reflection is shown by four rays marked 1, 2, 3 and 4. Which of the four rays correctly shows the direction of reflected ray?</p>
<div style='float: left; width: 50%'>
<div class="fragment">
<p>(a) 1</p>
<p>(b) 2</p>
<p>(c) 3</p>
<p>(d) 4</p>
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<img src="https://cdn.mathpix.com/cropped/2024_01_24_b3449e08884b7d728783g-153.jpg?height=355&width=376&top_left_y=916&top_left_x=620" height="200px">
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<p><strong>Ques.</strong> The optical density of turpentine is higher than that of water while its mass density is lower. Figure shows a layer of turpentine floating over water in a container. For which one of the four rays incident on turpentine in figure, the path shown is correct?</p>
<div style='float: left; width: 50%'>
<div class="fragment">
<p>(a) 1</p>
<p>(b) 2</p>
<p>(c) 3</p>
<p>(d) 4</p>
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<img src="https://cdn.mathpix.com/cropped/2024_01_24_b3449e08884b7d728783g-153.jpg?height=295&width=415&top_left_y=1703&top_left_x=544" height="180px">
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<p><strong>Ques.</strong> A car is moving with at a constant speed of 60 km h$^{-1}$ on a straight road. Looking at the rear view mirror, the driver finds that the car following him is at a distance of 100 m and is approaching with a speed of 5 kmh$^{-1}$. In order to keep track of the car in the rear, the driver begins to glance alternatively at the rear and side mirror of his car after every 2 s till the other car overtakes. If the two cars were maintaining their speeds, which of the following statement (s) is/are correct?</p>
<div class="fragment">
<p>(a) The speed of the car in the rear is 65 km h$^{-1}$</p>
<p>(b) In the side mirror, the car in the rear would appear to approach with a speed of 5 kmh$^{-1}$ to the driver of the leading car</p>
<p>(c) In the rear view mirror, the speed of the approaching car would appear to decrease as the distance between the cars decreases</p>
<p>(d) In the side mirror, the speed of the approaching car would appear to increase as the distance between the cars decreases</p>
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<p><strong>Ques.</strong> Consider an extended object immersed in water contained in a plane trough. When seen from close to the edge of the trough the object looks distorted because</p>
<div class="fragment">
<p>(a) the apparent depth of the points close to the edge are nearer the surface of the water compared to the points away from the edge</p>
<p>(b) the angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air</p>
<p>(c) some of the points of the object far away from the edge may not be visible because of total internal reflection</p>
<p>(d) water in a trough acts as a lens and magnifies the object</p>
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<p><strong>Thinking process :</strong></p>
<p>This problem is based on the phenomenon of reflection when it goes from denser to rare medium</p>
<p><strong>Ans.</strong> (a, b, c)</p>
<p>When immersed object is seen from close to the edge of the trough the object looks distorted because the apparent depth of the points close to the edge are nearer the surface of the water compared to the points away from the edge.</p>
<p>The angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air and some of the points of the object far away from the edge may not be visible because of total internal reflection.</p>
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<p><strong>Ques.</strong> A rectangular block of glass ABCD has a refractive index 1.6. A pin is placed midway on the face AB figure. When observed from the face AD, the pin shall</p>
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<div class="fragment">
<p>(a) appear to be near A</p>
<p>(b) appear to be near D</p>
<p>(c) appear to be at the centre of AD</p>
<p>(d) not be seen at all</p>
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<img src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/17011f23-0f58-4033-16c3-86c51d26b000/public" height="160px">
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<p><strong>Ans.</strong> (a, d)</p>
<p>Alexandar’s dark band lies between the primary and secondary rainbows, forms due to light scattered into this region interfere destructively.</p>
<p>Since, primary rainbows subtends an angle nearly $41^{\circ}$ to $42^{\circ}$ at observer’s eye, whereas, secondary rainbows subtends an angle nearly $51^{\circ}$ to $54^{\circ}$ at observer’s eye w.r.t. incident light ray.</p>
<p>So, the scattered rays with respect to the incident light of the sun lies between approximately $42^{\circ}$ and $50^{\circ}$.</p>
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<p><strong>Thinking process :</strong></p>
<p>The magnifying power $m$ is the ratio of the angle $\beta$ subtended at the eye by the final image to the angle ${\alpha}$ which the object subtends at the lens or the eye. Hence, in normal adjustment</p>
<p>$m \approx \frac{\beta}{\alpha} \approx \frac{h}{f_e} \frac{f_o}{h}=\frac{f_o}{f_e}$</p>
<p>In this case, the length of the telescope tube is $f_o+f_e$.</p>
<p><strong>Ans.</strong> (a, b, c)</p>
<p>The length of the telescope tube is</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Thinking process :</strong></p>
<p>The length of image is the separation between the images formed by mirror of the extremities of object.</p>
<p><strong>Ans.</strong></p>
<p>Since, the object distance is $u$. Let us consider the two ends of the object be at distance $u_1=u-\frac{L}{2}$ and $u_2=u+\frac{L}{2}$, respectively so that $|u_1-u_2|=L$. Let the image of the two ends be formed at $v_1$ and $v_2$, respectively so that the image length would be</p>
<p>$L^{\prime}=|v_1-v_2|$</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Ques.</strong> A circular disc of radius $R$ is placed co-axially and horizontally inside an opaque hemispherical bowl of radius ‘$a$’ figure. The far edge of the disc is just visible when viewed from the edge of the bowl. The bowl is filled with transparent liquid of refractive index $\mu$ and the near edge of the disc becomes just visible. How far below the top of the bowl is the disc placed?</p>
<img src="https://cdn.mathpix.com/cropped/2024_01_24_b3449e08884b7d728783g-159.jpg?height=271&width=518&top_left_y=1698&top_left_x=520" height="170px">
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<p><strong>Thinking process :</strong></p>
<p>This problem involves the wide application of geometry and Snell’s law.</p>
<p><strong>Ans.</strong></p>
<p>Refering to the figure, $A M$ is the direction of incidence ray before liquid is filled. After liquid is filled in, $B M$ is the direction of the incident ray. Refracted ray in both cases is same as that along $A M$.</p>
<p>Let the disc is separated by $O$ at a distance $d$ as shown in figure. Also, considering angle</p>
<p>Here, in figure</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Ans.</strong></p>
<p>If there was no cut, then the object would have been at a height of $0.5 cm$ from the principal axis $OO^{\prime}$.</p>
<p>Applying lens formula, we have</p>
<p>$ \frac{1}{v}-\frac{1}{u}  =\frac{1}{f} $</p>
<p>$ \therefore  \frac{1}{v}  =\frac{1}{u}+\frac{1}{f}=\frac{1}{-50}+\frac{1}{25}=\frac{1}{50} $</p>
<p>Mangnification is m = $\frac{v}{u}=-\frac{50}{50}=-1 $</p>
<p>Thus, the image would have been formed at 50 cm from the pole and 0.5 cm below the principal axis. Hence, with respect to the $X$-axis passing through the edge of the cut lens, the coordinates of the image are (50 cm, -1 cm).</p>
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<p><strong>Thinking process :</strong></p>
<p>This is also one of the methods for finding the focal length of the lens in laboratory and known as displacement method.</p>
<p><strong>Ans.</strong></p>
<p>Principal of reversibility is states that the position of object and image are interchangeable. So, by the versibility of $u$ and $v$, as seen from the formula for lens.</p>
<p>$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$</p>
<p>It is clear that there are two positions for which there shall be an image.</p>
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Ans.</strong></p>
<p>Let us consider a portion of a ray between $x$ and $x+d x$ inside the liquid. Let the angle of incidence at $x$ be $\theta$ and let it enter the thin column at height $y$. Because of the bending it shall emerge at $x+d x$ with an angle $\theta+d \theta$ and at a height $y+d y$. From Snell’s law,</p>
<img src="https://cdn.mathpix.com/cropped/2024_01_24_b3449e08884b7d728783g-164.jpg?height=344&width=418&top_left_y=1884&top_left_x=660" height="150px">
<p><strong><Success>Sol. to be continued …</Success></strong></p>
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<p><strong>Ques.</strong> If light passes near a massive object, the gravitational interaction causes a bending of the ray. This can be thought of as happening due to a change in the effective refractive index of the medium given by</p>
<p>$n(r)=1+2 G M / r c^{2}$</p>
<p>where $r$ is the distance of the point of consideration from the centre of the mass of the massive body, $G$ is the universal gravitational constant, $M$ the mass of the body and $c$ the speed of light in vacuum. Considering a spherical object find the deviation of the ray from the original path as it grazes the object.</p>
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<p><strong>Ques.</strong>  An infinitely long cylinder of radius $R$ is made of an unusual exotic material with refractive index-1 (figure). The cylinder is placed between two planes whose normals are along the $y$-direction. The centre of the cylinder 0 lies along the $y$-axis. A narrow laser beam is directed along the $y$-direction from the lower plate. The laser source is at a horizontal distance $x$ from the diameter in the $y$ direction. Find the range of $x$ such that light emitted from the lower plane does not reach the upper plane.</p>
<img src="https://cdn.mathpix.com/cropped/2024_01_24_b3449e08884b7d728783g-166.jpg?height=386&width=529&top_left_y=1582&top_left_x=546" height="170px">
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<p><strong>Ques.</strong> (i) Consider a thin lens placed between a source (S) and an observer (O). Let the thickness of the lens vary as $w(b)=w_0-\frac{b^{2}}{\alpha}$, where $b$ is the verticle distance from the pole, $w_0$ is a constant. Using Fermat’s principle i.e., the time of transit for a ray between the source and observer is an extremum find the condition that all paraxial rays starting from the source will converge at a point 0 on the axis. Find the focal length.</p>
<img src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/8c420667-6526-4cce-5b91-029d31ca1000/public" height = "120px"/>
<p><strong><Success>Ques. to be continued …</Success></strong></p>
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